Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

While doing revision, I came across this problem:

The surface given by $z=x^2-y^2$ is cut by the plane given by $y=3x$, producing a curve in the plane. Find the slope of this curve at the point $(1,3,-8)$.

I tried substituting $y=3x$ into $z=x^2-y^2$, yielding $z=-8x^2$. Then, $\frac{dz}{dx}=-16x=-16$.

However the answer is $-8\sqrt{\frac{2}{5}}$.

Thank you very much for any help.

share|cite|improve this question
What do you mean by slope of a line in dimension 3? – Mercy King Jun 23 '12 at 11:02
@Mercy I think the question requires us to consider the shape of the intersection in the plane $y=3x$. Re orientation I suppose we should regard $z$ as the dependent variable. – hwhm Jun 23 '12 at 11:14
According to; It is meaningless to talk about the slope of a curve in three-dimensional space unless the slope with respect to what is specified. – Babak S. Jun 23 '12 at 11:18
This curve can be parametrized by $\gamma(t)=(t,3t,-8t^2)$ so I don't really understand the notion of slope here! – Mercy King Jun 23 '12 at 11:21
@Mercy You might wanna do a 3D plot. The question is badly posed but it's clear from the plot that what's wanted is to have axes introduced to the plane $y=3x$ (I'd say the horizontal axes is along $(1,3,0)$ and the vertical along $(0,0,1)$) and then talk about the intersection as though it is graphed on that plane. – hwhm Jun 23 '12 at 11:25
up vote 3 down vote accepted

This is a very badly posed question, and does not have an answer. (Read the comments.)

The following is a solution to a rephrased question which can be answered, however.

Question: The surface given by $z=x^2−y^2$ is cut by the plane given by $y=3x$, producing a curve in the plane.

Treating the intersection as a curve in the said plane with vertical axis along $(0,0,1)$ and horizontal axis along $(1,3,0)/\sqrt{10}$, find the slope of this curve at the point (1,3,−8).

Solution: Any point on the plane has Cartesian coordinates in the form $$\frac{a}{\sqrt{10}}\begin{pmatrix} 1\\3 \\0 \end{pmatrix} + b\begin{pmatrix} 0\\0\\1 \end{pmatrix}.$$ Substituting this into $z=x^2−y^2$, we get $b = -4a^2/5$.

So the "slope" at a point on this intersection, with $a$ and $b$ given, is $$\frac{db}{da} = -\frac{8a}{5}.$$

Setting $$\begin{pmatrix}1 \\3\\-8\end{pmatrix} = \frac{a}{\sqrt{10}}\begin{pmatrix} 1\\3 \\0 \end{pmatrix} + b\begin{pmatrix} 0\\0\\1 \end{pmatrix},$$ we get $a = \sqrt{10}$ and so the "slope" at this point is $-8\sqrt{\frac{2}{5}}$.

Solution using grad: Let $f:=x^2-y^2-z$ and $g:=y-3x$. At the point $(1,3,-8)$, $\nabla f=(2,-6,-1)$ and $\nabla g=(-3,1,0)$. Their cross product, $(1,3,-16)$, is along the tangent direction of the intersecting curve produced by the surface and the plane, at the point $(1,3,-8)$.

Denote the angle between $(1,3,-16)$ and $(1,3,0)$ (i.e. the "horizontal") by $\theta$. Then, using the dot product, $\cos\theta = \sqrt{\frac{5}{133}}$. The "slope" is $$\tan \theta = - \sqrt{\frac{1}{\cos^2 \theta}-1} = -8\sqrt{\frac{2}{5}},$$ where the negative square root is taken because the "vertical" is along $(0,0,1)$.

share|cite|improve this answer
Thanks a lot. Is there any way to solve this question using the Gradient operator? ($\nabla F$) – yoyostein Jun 24 '12 at 7:13
@yoyostein Of course - see the updated answer. I didn't know the question was meant to be undergrad level then - it did look like a very badly phrased high school level question! – hwhm Jun 25 '12 at 3:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.